CBSE Class 09 Mathematics
Revison Notes
CHAPTER 12
HERON’S FORMULA

1. Area of a Triangle – by Heron’s Formula

2. Application of Heron’s Formula in finding Areas of Quadrilaterals

• Triangle with base "b" and altitude "h" is

Area =12×(b×h)Area =12×(b×h)

• Area of an isosceles triangle whose equal side is aa = a22a22 square units
• Triangle with sides a, b and c

(i) Semi perimeter of triangle s =a+b+c2a+b+c2

(ii) Area=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√s(s−a)(s−b)(s−c)sq. unit

• Equilateral triangle with side "a"

Perimeter = 3a3a units

Altitude = 3√2a32a units

Area =3√4a2 square unitsArea =34a2 square units

• Rectangle with length ll, breadth bb 

Perimeter = 2(l+b)2(l+b)

Area = l×bl×b

• Square with side aa

Perimeter = 4a4a units

Area = a2a2  sq. units

Area = (Diagonal)2(Diagonal)2 sq. units

• Parallelogra with length ll, breadth bb and height hh

Perimeter = 2(l+b)2(l+b)

Area = b×hb×h

• Trapezium with parallel sides "a" & "b" and the distance between two parallel sides as "h".

Area =12(a+b)hsquare unitsArea =12(a+b)hsquare units